Spatial extent of an outbreak in animal epidemics In collaboration - PowerPoint PPT Presentation

Spatial extent of an outbreak in animal epidemics In collaboration with E. Dumonteil, S. N. Majumdar, A. Zoia O PNAS 110, 4239 (2013)

SIR model for epidemics Three species : susceptibles (S), infected (I), recovered (R) dS dt = − β I S • mean field fully connected model dI • β rate of infection transmission dt = β I S − γ I dR • γ rate at which an infected recovers dt = γ I I ( t ) + S ( t ) + R ( t ) = N N being the total population

Outbreak of an epidemic Initial condition : I (0) = 1, S (0) = N − 1 ≈ N , R (0) = 0 Outbreak regime dS dt = − β I S t ≈ 0 , S ≈ N dI dI dt ' ( β N � γ ) I dt = β I S − γ I dR dt = γ I Reproduction rate: R 0 = β N γ

Deterministic and stochastic models SIR is a deterministic model. In the outbreak fluctuations are important • Stochastic process: Galton-Watson (mean field) • each infected individual transmits the disease at rate N β • each infected individual recovers at rate γ • R 0 < 1 epidemics extinction • R 0 > 1 epidemics invasion Reproduction rate: R 0 = β N γ • R 0 = 1 critical case

How far in space can an epidemic spread? Problem 1 : How to model the space? O The good candidate: Brownian process with branching and death In dt , each infected can: • recovers with probability γ dt • infects with probability β N dt = γ R 0 dt • otherwiese, it di ff uses ( D di ff usion const.)

Problem 2 : How to quantify the area that needs to be quarantined? CONVEX HULL Algorithms: Graham Scan (Nlog(N))

Monitoring the outbreak Day 1 Day 2 Day 3 O O O y y y x x x

Real applications Eric Dumonteuil

How to compute the convex hull of Branching processes? y C Cauchy formulas M( θ ) θ x O Z 2 π L = M ( θ ) d θ 0 Z 2 π A = 1 M 2 ( θ ) − ( M 0 ( θ )) 2 ⇤ ⇥ d θ 2 0

Support Function M ( θ ) = max 0 ≤ τ ≤ t [ x τ cos θ + y τ sin θ ] y x y x m 0 y ( t m ) y ( t m ) O 0 0 x t t x m t m 0 0 t m 0 • x m ( t ) x -maximum up to time t M (0) = x τ = t m = x m ( t ) • t m time location of the maximum M 0 ( θ = 0) = − x t m sin θ + y t m cos θ | θ =0 = y t m

y x y x m 0 y ( t m ) y ( t m ) O 0 0 x t t x m t m 0 0 t m 0 h L ( t ) i = 2 π h x m ( t ) i h x 2 m ( t ) i � h y 2 ( t m ) i ⇥ ⇤ h A ( t ) i = π consider a 1d branching process evolving in (0 , t ) • x m is the global maximum • t m is the location of the maximum • h y 2 ( t m ) i = . . . = 2 D t m

Backward Fokker Planck equation Q t ( x m ) = Proba[global max up to t < x m ] Q t + dt ( x m ) = γ dt + R 0 γ dtQ 2 t ( x m ) + [1 � γ ( R 0 + 1)] dt h Q t ( x m � ∆ x ) i t ( x m ) + h ∆ x 2 • h Q t ( x m � ∆ x ) i = Q t ( x m ) � h ∆ x i Q 0 2 i Q 00 t ( x m ) + . . . • h ∆ x i = 0 • h ∆ x 2 i = 2 Ddt h Q t ( x m � ∆ x ) i = Q t ( x m ) + Ddt ∂ 2 x Q t ( x m ) + . . .

∂ tQ = D ∂ 2 ∂ Q − γ ( R 0 + 1) Q + γ R 0 Q 2 + γ ∂ x 2 m • initial condition Q t =0 ( x m ) = θ ( x m ) • boundary condition Q t ( x m < 0) = 0 • boundary condition Q t ( x m → ∞ ) = 1 Z ∞ h L ( t ) i = 2 π [1 � Q t ( x m )] dx m . 0

Solid lines: our predictions 10 -2 R 0 =1 R 0 =1.15 1 100 0 . 1 = R 0 R 0 =0.99 10 -3 80 Prob(L,t) Blue lines: super-critical < L(t) > 60 R 0 =0.85 10 -4 40 Red lines: critical 10 -5 20 10 -6 0 10 1 10 2 10 3 10 4 10 5 10 1 10 2 Green lines: sub-critical t L 3500 10 -2 R 0 =1.01 1 R 0 =1.15 = 3000 R 0 10 -3 2500 Symbols: Monte Carlo 10 -4 Prob(A,t) < A(t) > 2000 10 -5 simulations R 0 =0.99 1500 10 -6 10 -7 1000 Dashed lines: analytical R 0 =0.85 10 -8 500 asymptotic results 10 -9 0 10 1 10 2 10 3 10 4 10 5 10 0 10 1 10 2 10 3 10 4

The critical case s 6 D γ + O ( t − 1 / 2 ) h L ( t ! 1 ) i = 2 π h A ( t ! 1 ) i = 24 π D ln t + O (1) 5 γ 3500 R 0 =1 1 R 0 =1 R 0 =1.15 R 0 =1.15 0 1 100 0 3000 . . 1 1 = = R 0 R 0 R 0 =0.99 2500 80 < A(t) > < L(t) > 2000 = 0 . 9 9 R 60 R 0 =0.85 0 1500 40 1000 R 0 =0.85 20 500 0 0 10 1 10 2 10 3 10 4 10 5 10 1 10 2 10 3 10 4 10 5 t t

When t → ∞ the perimeter remains finite, but the area diverges! How it is possible ? ... Fluctuations 24 π D t = ∞ t = ∞ A − 2 L →∞ L − 3 Prob( A ) Prob( L ) − − − − → − − − − → 5 γ A →∞ 10 -2 10 -2 10 -3 10 -3 10 -4 Prob(A,t) Prob(L,t) 10 -5 10 -4 10 -6 10 -7 10 -5 10 -8 10 -9 10 -6 10 0 10 1 10 2 10 3 10 4 10 5 10 6 10 1 10 2 10 3 A L

Out of criticality When R 0 6 = 1, characteristic time t ∗ ⇠ | R 0 � 1 | − 1 . For times t < t ∗ the epidemic behaves as in the critical regime. In the subcritical regime, for t > t ∗ the epidemic goes to extinction. In the supercritical regime, with probability 1 � 1 /R 0 epidemic explodes. 3500 R 0 =1 1 R 0 =1 R 0 =1.15 R 0 =1.15 0 1 100 0 3000 . . 1 1 = = R 0 R 0 R 0 =0.99 2500 80 < A(t) > < L(t) > 2000 = 0 . 9 9 R 60 R 0 =0.85 0 1500 40 1000 R 0 =0.85 20 500 0 0 10 1 10 2 10 3 10 4 10 5 10 1 10 2 10 3 10 4 10 5 t t

Supercritical ✓ ◆ p 1 � 1 h L ( t � t ∗ ) i = 4 π D γ ( R 0 � 1) t R 0 ✓ 1 � 1 ◆ D γ ( R 0 � 1) t 2 h A ( t � t ∗ ) i = 4 π R 0 t ∗ ⇠ | R 0 � 1 | − 1 10 5 10 4 5 5 . 5 2 2 . = 2 . 1 R 0 =1.25 = 5 = R 0 10 2 R 0 =1.5 . R 0 1 10 3 R 0 = < A(t) > < L(t) > R 0 1 = R 0 R 0 =1 10 2 10 1 10 1 10 0 10 0 10 1 10 2 10 3 10 0 10 1 10 2 10 3 t t

∂ tW = D ∂ 2 ∂ W + γ ( R 0 − 1) W − γ R 0 W 2 ∂ x 2 m 1 t = 2 0 0 t=400 t = ∞ x 0 10 30 40 1 − 1 ∞ t = R 0 t=400 t=100 t=200 0 0 0 20 40 60 80 100 120 140 x Traveling front solution

Conclusions: • Branching Brownian motion with death as a model for the spatial extent of animal epidemics • Using Cauchy Formulas we can map the convex hull problem in the extreme statistic of the 1- dimensional process • Backward F-P equations for the extreme distributions • Critical case has very large fluctuations • Super Critical case: traveling front solution

How far in space can an epidemic spread? Problem 1 : How to model the space? The population is uniformly distributed At time t = 0 an infected individual appears ... and moves in space Brownian motion is the paradigm of animal migration while human beings take the plane (even when they are sick)

Similar calculations allows to express the mean area as: Z ∞ h A ( t ) i = π dx m [2 x m (1 � Q t ( x m )) � T t ( x m )] 0 Where the evolution of T t ( x m ) is governed by: D ∂ 2  � ∂ ∂ tT t + ∂ x Q t ( x m ) = + 2 γ R 0 Q t − γ ( R 0 + 1) T t , ∂ x 2 m Both PDE can be integrated numerically and solved in some asymptotic limit

Dimensional reduction Z 2 π L = M ( θ ) d θ 0 Z 2 π A = 1 M 2 ( θ ) − ( M 0 ( θ )) 2 ⇤ ⇥ d θ 2 0 If the process is rotationally invariant any average is independent of θ h L ( t ) i = 2 π h M (0) i h M 2 (0) i � h M 0 (0) 2 i ⇥ ⇤ h A ( t ) i = π h L ( t ) i = 2 π h x m ( t ) i h x 2 m ( t ) i � h y 2 ( t m ) i ⇥ ⇤ h A ( t ) i = π This relation is valid ONLY in average